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About me

About Me:
I am a mathematics student at Warwick. I pride myself on having a deep and thorough understanding of mathematical concepts from A level to below. I feel I have encountered the most common classic mistakes mathematics students make as I have made them myself. Throughout my mathematics career I have made sure that I rectify any small misunderstanding I have and I feel this puts me in a great position to teach others about the subject I love.
I have privately tutored students at mathematics GCSE and A level and I have also volunteered for a year in primary school mathematics lesson as a classroom assistant. I believe everyone can be great at mathematics and that any self doubt is usually due to small fixable mistakes in the understanding of basic concepts.
The Sessions:
I will focus primarily on any particular questions you may have for me. If we get through these questions I like to work on exam style questions as this is the best way to improve your grade. If it becomes clear that there is a fundamental gap n your knowledge on a certain topic then we will take a step back and slowly go through the theory of that topic.
For this style of tutoring the exam board you are on is very important so the sooner I could get this information the better.
Oxford, MAT, STEP:
I have personally had interviews for Oxford for a mathematics degree and will try my best to advise you on interview technique and share my experience. I have also achieved a 2 in STEP 1 and would be willing to look over any questions you might have about it, although I must stress, unlike A Level, my knowledge on STEP is far from comprehensive. I also passed the MAT and would be wiling to advise on that front.
Questions:
For any questions send me a 'WebMail' or book a 'Meet the Tutor Session'! (both accessible through this website).

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Questions Hugh has answered

For what values of k does the graph y=x^(2)+2kx+5 not intersect the x-axis

Where the graph intersects the x-axis, x2+2kx+5 must be equal to zero. Thus we
can answer the equivalent question: For what k does x2+2kx+5 = 0 not have a
solution?
This is now a simpler problem (roots of a quadratic equation). We can apply
the common method of considering the discriminant of...

Where the graph intersects the x-axis, x2+2kx+5 must be equal to zero. Thus we can answer the equivalent question: For what k does x2+2kx+5 = 0 not have a solution?

This is now a simpler problem (roots of a quadratic equation). We can apply the common method of considering the discriminant of x2+2kx+5. Using standard quadratic formula notation where in this case a=1, b =2k and c=5 we evaluate the discriminant : b2-4ac= (2k)2-4*1*5 = 4k2 -20.

Now since the discriminant appears in a square root sign in the quadratic equation, if it is negative there can be no real solutions to the equation ( great this is what we want!).

Thus we want discriminant negative: 4k2 -20 <0. Divide both sides of the inequality by 4 so we have k2-5<0.

Now this is where we must take great care, the following reasoning is a common MISTAKE: rearragne the inequality so we have k2 < 5, then squarrot both sides so we have k < sqrt(5) or k < - sqrt(5) . The second inequalit is implied by the first thus the discriminate negtive for all k values les then the sqrt(5). THIS IS INCORRECT.

When dealing with inequalities involving powers such as we are here we must be extremely careful. the mistake in the reasoning above is when we say k < - sqrt(5), this is actually a form of the common mistake of not inverting the inequality when multiplying both sides of an equation by a negative. Instead when dealing with inequalities with powers it is always much wiser to sketch a graph of the situation.

k2 - 5 is the standard quadratic U shape (think y=x2) shifted down by 5. Having sketched this out it is clear that this graph is less then 0 when it is inbetween it's two roots.

If cos(x)= 1/3 and x is acute, then find tan(x).

Consider a right angled triangle. Call one of the angles (not the right angle)
in this triangle x. We can do this as we are told x is acute. The side opposite
to x label O, the side adjacent to x label A, and label the hypotenuse H.
Now from SOHCAHTOA cos(x) = A/H = 1/3 and tan(x) = O/A . We a...

Consider a right angled triangle. Call one of the angles (not the right angle) in this triangle x. We can do this as we are told x is acute. The side opposite to x label O, the side adjacent to x label A, and label the hypotenuse H.

Now from SOHCAHTOA cos(x) = A/H = 1/3 and tan(x) = O/A . We also know by pythagoras that A2 + O2 = H2 . We shall now combine these equations to get our result.

A/H = 1/3 implies H = 3A implies H2 = 9A2. Substituting this result into our euation obtained by pythagoras we get: A2 + O2 = 9A2. Rearranging: O2 = 8A2 implies O2/A2 = 8 implies (O/A)2 = 8. Now we take the square root of both sides. Here we must take care, O and A are lengths and so are not negative, so we only consider the positive root: O/A = sqrt(8) = tan(x) and so we are done.

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